Application of L'Hôpital's rule to a definition of $\mathrm{e}$

Question

$$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^ n $$ So the problem above is giving me trouble, the answer key I was giving says it equals to $\mathrm{e}$, but I have no idea how to get a answer. I look at the problem and plug in infinity to see if the indeterminate form appears, but I cannot tell, since I get $$\left(1+\frac{1}{\infty}\right)^\infty$$ I guess the problem here is what does $$\frac{1}{\infty}$$ equal? I think it equals infinity. If that is true, I get the form $$\infty^\infty$$ so now I use L.H. but no matter the derivative I get I have no clue how the answer can possibly be $\mathrm{e}$.

Answer 1

In many contexts, this is the very definition of $e$.

However if you want:

$$\begin{align*} \lim_{n\to\infty}(1+\frac1n)^n&=\lim_{n\to\infty}\exp\left(n\ln(1+\frac1n)\right)\\ &=\lim_{n\to\infty} \exp\left(\frac{\ln(1+\frac1n)}{n^{-1}}\right)\\ &=\exp\left(\lim_{n\to\infty}\frac{\ln(1+\frac1n)}{n^{-1}} \right)\\ &\stackrel{LR}{=}\exp\left(\lim_{n\to\infty}\frac{\frac{-n^{-2}}{1+1/n}}{-n^{-2}} \right)\\ &=\exp\left(1\right)\\ &=e \end{align*}$$

Answer 2

The indeterminate form is actually $(1+\frac{1}{\infty})^{\infty} =(\frac{\infty}{\infty})^{\infty} $.

With regards to the limit, there are many proofs here that show that $(1+\frac1{n})^n $ is increasing and bounded and so must have a limit.

Here is one of them:

Proving : $ \bigl(1+\frac{1}{n+1}\bigr)^{n+1} \gt (1+\frac{1}{n})^{n} $

Answer 3

If you define $e$ as $\sum_{k=0}^{\infty}{1\over k!}={1\over 1}+{1\over 1}+{1\over 2}+{1\over 6}...$ then you have $$(1+\frac1n)^n=\sum_{k=0}^{n}{n\choose k}{1\over n^k}=\sum_{k=0}^{n}{n(n-1)(n-2)...(n-k+1)\over k!n^k}=\sum_{k=0}^{n}{n^k+c_1n^{k-1}+...c_kn^0\over k!n^k}$$

Therefore,$$\lim_{n\to \infty}(1+\frac1n)^n=\lim_{n\to \infty}\sum_{k=0}^{n}{n^k+c_1n^{k-1}+...c_kn^0\over k!n^k}=\sum_{k=0}^{\infty}{1\over k!}=e$$

Answer 4

Hint

Using L'Hospital rule to $$A_n=(1+\frac{1}{n})^ n$$ take logarithms $$\log(A_n)=n\log(1+\frac{1}{n})=\frac{\log(1+\frac{1}{n})}{\frac 1n}$$ Make now $x=\frac 1n$ which makes $$\log(A_n)=\frac{\log(1+x)}x$$ Now, apply the rule $(x\to 0)$

Answer 5

For $n\in N$ we have $$1/(n+1)=\int_n^{n+1}1/(n+1)\;dx<\int_n^{n+1}(1/x)\;dx <\int_n^{n+1}1/n\;dx=1/n,$$ $$\text {and }\quad \int_n^{n+1}(1/x)\; dx=\ln (n+1)-\ln n=\ln (1+1/n).$$ $$\text {Therefore }\quad 1/(n+1)<\ln (1+1/n) <1/n$$ $$\text { so }\quad n/(n+1)<\log (1+1/n)^n<1$$ $$\text {so }\quad e\cdot e^{-1/(n+1)}<(1+1/n)^n<e.$$

Remark: Apply this method to $\ln (1+y/z)=\ln (y+z)-\ln z=\int_z^{y+z}(1/x)\;dx$ for any $y\in R$ and for $\min (1,1-y)<z\in R$ and you obtain $\lim_{z\to \infty}(1+y/z)^z=e^y.$